Reverse Triangle Inequality/Real and Complex Fields/Proof 2

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Theorem

Let $x$ and $y$ be elements of either the real numbers $\R$ or the complex numbers $\C$.

Then:

$\cmod {x - y} \ge \size {\cmod x - \cmod y}$


Proof

From the proof 2 of the corollary to this result, which is derived independently:

$\size {x - y} \ge \size x - \size y$


There are two cases:

$(1): \quad \size x \ge \size y$

We have :

$\size {\size x - \size y} = \size x - \size y$

and the proof is finished.

$\Box$


$(2): \quad \size y \ge \size x$

We have:

$\size {y - x} \ge \size y - \size x = \size {\size y - \size x}$

But:

$\size {y - x} = \size {x - y}$

and:

$\size {\size y - \size x} = \size {\size x - \size y}$


From this we have:

$-\size {\size x - \size y} \ge -\size {x - y}$

Since, by Negative of Absolute Value, we have that:

$\size x - \size y \ge -\size {\size x - \size y}$

it follows that:

$-\size {x - y} \le \size x - \size y \le \size {x - y}$

The result follows.

$\blacksquare$


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